An electron, placed in an electric field, experiences a force F of $1\, N.$ What are the magnitude and direction of the electric field E at the point where the electron is located $(e = 1.6 \times 10^{-19} C)$ ?
- $\frac {1}{e} N/C, F$ and E are along the same direction
- $\frac {1}{e} N/C, F$ and E are against each other
- $\frac {1}{e} N/C, F$ and E are perpendicular
- $N/C, F$ and E are against each other
The Correct Option is B
Solution and Explanation
$E=\frac{F}{q}$
where, $F$ is the magnitude of the electrostatic force and $q$ is the charge
Here, $F=1\, N$
and $q=e=1.6 \times 10^{-19} C $
$\Rightarrow E=\frac{1}{e} N / C$
Force experienced by a negative charge is in the opposite direction to the electric field. As, for a negatively charged particle, its electric field vector at each point is directed radially inwards.
Thus, the magnitude and direction of $E$ is $\frac{1}{e} N / C$ and $E$ and $F$ are against each other.
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Concepts Used:
Electric Field
Electric Field is the electric force experienced by a unit charge.
The electric force is calculated using the coulomb's law, whose formula is:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).